Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days and a standard deviation of 15 days.

a. In what range would you expect to find the middle 95% of most pregnancies?

b. If you were to draw samples of size 43 from this population, in what range would you expect to find the middle 95% of most averages for the lengths of pregnancies in the sample?

c. In what range would you expect to find the middle 50% of most pregnancies?

d. If you were to draw samples of size 36 from this population, in what range would you expect to find the middle 50% of most averages for the lengths of pregnancies in the sample?

Answer 1

(a)

= 265

= 15

Middle 95% corresponds to area = 0.95/2 = 0.475 from mid value on either side of mid value.

Table of Area Under Standard Normal Curve gives Z = 1.96

Low side:
Z = - 1.96 = (X - 265)/15

X = 265 - (1.96 X 15) = 265 - 29.4 = 235.6

X = 265 + (1.96 X 15) = 265 + 29.4 = 294.4

So,

the range is given by:

235.6 to 294.4

(b)

= 265

= 15

n = 43

SE = /

= 15/

= 2.2875

Middle 95% corresponds to area = 0.95/2 = 0.475 from mid value on either side of mid value.

Table of Area Under Standard Normal Curve gives Z = 1.96

Low side:
Z = - 1.96 = (X - 265)/2.2875

X = 265 - (1.96 X 2.2875) = 265 - 4.4835 = 260.5165

X = 265 + (1.96 X 2.2875) = 265 + 4.4865 = 269.4865

So,

the range is given by:

260.5165 to 269.4865

(c)

= 265

= 15

Middle 50% corresponds to area = 0.50/2 = 0.25 from mid value on either side of mid value.

Table of Area Under Standard Normal Curve gives Z = 1.675

Low side:
Z = - 1.675 = (X - 265)/15

X = 265 - (1.675 X 15) = 265 - 25.125 = 239.875

X = 265 + (1.675 X 15) = 265 + 25.125 = 290.125

So,

the range is given by:

239.875 to 290.125

(d)

= 265

= 15

n = 36

SE = /

= 15/

= 2.50

Middle 50% corresponds to area = 0.50/2 = 0.25 from mid value on either side of mid value.

Table of Area Under Standard Normal Curve gives Z = 1.675

Low side:
Z = - 1.675 = (X - 265)/2.50

X = 265 - (1.675 X 2.50) = 265 - 4.1875 = 260.8125

X = 265 + (1.675 X 2.50) = 265 + 4.1875 = 269.1875

So,

the range is given by:

260.8125 to 269.1875